Solve the differential equation $y(xy^{''}+y^{'})=xy^{'2}(1-x)$ Solve the differential equation $y(xy^{''}+y^{'})=xy^{'2}(1-x)$
This is the integrable differential which can be solved by substitution $y^{'}=yz,z=z(x)$
Applying substitution gives $y(x(z+yz^{'})+yz)=y(xz+xzy^{'}+yz)\Rightarrow xz+xyz^{'}+yz-xyz^2+x^2yz^2=0$.
How to evaluate $y$ in terms of $z$?
 A: $$y(x)\left(xy''(x)+y'(x)\right)=xy'(x)^2\left(1-x\right)\Longleftrightarrow$$
$$\frac{y(x)\left(xy''(x)+y'(x)\right)}{y'(x)^2}=x\left(1-x\right)\Longleftrightarrow$$
$$\frac{y(x)}{y'(x)}+\frac{xy''(x)}{y'(x)^2}=x\left(1-x\right)\Longleftrightarrow$$
$$\frac{x^2y'(x)^2-xy'(x)+y'(x)y(x)+xy''(x)y(x)}{y'(x)^2}=0\Longleftrightarrow$$
$$x^2y'(x)^2-xy'(x)^2+y'(x)y(x)+xy''(x)y(x)=0\Longleftrightarrow$$

The equation is a homogeneous polynomial of degree 2 with respect to $y$,$y'(x)$ and $y''(x)$.
Letting $y(x)=e^{v(x)}$ will reduce the order of the equation.
This gives $y'(x)=e^{v(x)}v'(x)$ and $y''(x)=e^{v(x)}\left(v''(x)+v'(x)^2\right)$:

$$e^{2v(x)}\left(x^2v'(x)^2+xv''(x)+v'(x)\right)=0\Longleftrightarrow$$

For $e^{2v(x)}=0$ does no solution exist.
Let $v'(x)=u(x)$ which gives $v''(x)=u'(x)$:

$$xu'(x)+u(x)=-x^2u(x)^2\Longleftrightarrow$$
$$-\frac{u'(x)}{u(x)^2}-\frac{1}{xu(x)}=x\Longleftrightarrow$$

Let $w(x)=\frac{1}{u(x)}$, which gives $w'(x)=-\frac{u'(x)}{u(x)^2}$:

$$w'(x)-\frac{w(x)}{x}=x\Longleftrightarrow$$

Let $\mu(x)=e^{\int-\frac{1}{x}\space\text{d}x}=\frac{1}{x}$:

$$\frac{w'(x)}{x}-\frac{w(x)}{x^2}=1\Longleftrightarrow$$

Substitute $-\frac{1}{x^2}=\frac{\text{d}}{\text{d}x}\left(\frac{1}{x}\right)$:

$$\frac{w'(x)}{x}+\frac{\text{d}}{\text{d}x}\left(\frac{1}{x}\right)w(x)=1\Longleftrightarrow$$
$$\frac{\text{d}}{\text{d}x}\left(\frac{w(x)}{x}\right)=1\Longleftrightarrow$$
$$\int\frac{\text{d}}{\text{d}x}\left(\frac{w(x)}{x}\right)\space\text{d}x=\int 1\space\text{d}x\Longleftrightarrow$$
$$\frac{w(x)}{x}=x+\text{C}\Longleftrightarrow$$
$$w(x)=x\left(x+\text{C}\right)\Longleftrightarrow$$
$$u(x)=\frac{1}{x^2+x\text{C}}\Longleftrightarrow$$
$$v'(x)=\frac{1}{x^2+x\text{C}}\Longleftrightarrow$$
$$v(x)=\frac{\ln(x)-\ln(x+\text{C})}{\text{C}}+\text{C}_2\Longleftrightarrow$$
$$y(x)=\text{C}_2e^{\frac{\ln(x)-\ln(x+\text{C})}{\text{C}}}$$
